Differential Equations
SECOND ORDER (homogeneous)
Graham S McDonald
A Tutorial Module for learning to solve 2nd
order (homogeneous) differential equations
● Table of contents
● Begin Tutorial
c 2004 g.s.mcdonald@salford.ac.uk
Table of contents
1.
2.
3.
4.
5.
Theory
Exercises
Answers
Solving quadratics
Tips on using solutions
Full worked solutions
Section 1: Theory
3
1. Theory
In this Tutorial, we will practise solving equations of the form:
d2 y
dy
+b
+ cy = 0.
dx2
dx
i.e. second order (the highest derivative is of second order),
linear (y and/or its derivatives are to degree one) with
constant coefficients (a, b and c are constants that may be zero).
a
There are no terms that are constants and no terms that are only
a function of x. If such terms were present, it would be conventional to collect them together on the right-hand-side of the equation. Here, we simply have a zero on the right-hand-side of the equals
sign and this type of ordinary differential equation (o.d.e.) is called
“homogeneous”.
Since the o.d.e. is second order, we expect the general solution to
have two arbitrary constants (these will be denoted A and B).
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Section 1: Theory
4
A trial solution of the form y = Aemx yields an “auxiliary equation”:
am2 + bm + c = 0,
that will have two roots (m1 and m2 ).
The general solution y of the o.d.e. is then constructed from the possible forms (y1 and y2 ) of the trial solution. The auxiliary equation
may have:
i)
real different roots,
m1 and m2 → y = y1 + y2 = Aem1 x + Bem2 x
or
ii)
real equal roots,
m1 = m2 → y = y1 + xy2 = (A + Bx)em1 x
or
iii)
complex roots,
p ± iq → y = y1 + y2 ≡ epx (A cos qx + B sin qx)
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Section 2: Exercises
5
2. Exercises
Find the general solution of the following equations. Where boundary
conditions are also given, derive the appropriate particular solution
too.
Click on Exercise links for full worked solutions (there are 16 exercises in total).
i
h
d2 y
dy
y 0 = dx
Notation: y 00 = dx
2,
Exercise 1. 2y 00 + 3y 0 − 2y = 0
Exercise 2. y 00 − 2y 0 + 2y = 0
Exercise 3. y 00 − 2y 0 + y = 0
● Theory ● Answers ● Solving quadratics ● Tips
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Section 2: Exercises
6
Exercise 4. y 00 = −4y
Exercise 5. y 00 = 4y
Exercise 6. 36y 00 − 36y 0 + 13y = 0
Exercise 7. 3y 00 + 2y 0 = 0
Exercise 8. 16y 00 − 8y 0 + y = 0
Exercise 9. y 00 + 4y 0 + 5y = 0 ;
y(0) = 0
and y 0 (0) = 2
Exercise 10. y 00 + 6y 0 + 13y = 0 ; y(0) = 2
and y 0 (0) = 1
Exercise 11. y 00 + 6y 0 + 9y = 0 ; y(0) = 1
and y 0 (0) = 2
● Theory ● Answers ● Solving quadratics ● Tips
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Section 2: Exercises
7
Exercise 12.
d2 y
dy
+
+y =0
dx2
dx
Exercise 13.
d2 y
dy
−6
+ 9y = 0
2
dτ
dτ
Exercise 14.
d2 y
dy
+7
+ 12y = 0
2
dτ
dτ
Exercise 15.
d2 x
dx
+5
+ 6x = 0
2
dτ
dτ
Exercise 16. 4
d2 x
dx
+8
+ 3x = 0
2
dτ
dτ
● Theory ● Answers ● Solving quadratics ● Tips
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Section 3: Answers
8
3. Answers
1
1. y = Ae 2 x + Be−2x ,
2. y = ex (A cos x + B sin x) ,
3. y = (A + Bx)ex ,
4. y = A cos 2x + B sin 2x ,
5. y = Ae2x + Be−2x ,
1
6. y = e 2 x (A cos x3 + B sin x2 ) ,
2
7. y = A + Be− 3 x ,
1
8. y = (A + Bx)e 4 x ,
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Section 3: Answers
9
9. y = e−2x (A cos x + B sin x);
y = 2e−2x sin x ,
10. y = e−3x (A cos 2x + B sin 2x);
11. y = (A + Bx)e−3x ;
1
12. y = e− 2 x (A cos
y = (1 + 5x)e−3x ,
√
3x
2
y = 12 e−3x (4 cos 2x + 7 sin 2x) ,
√
+ B sin
3
2 x)
,
13. y = (A + Bτ )e3τ ,
14. y = Ae−4τ + Be−3τ ,
15. x = Ae−2τ + Be−3τ ,
3
1
16. x = Ae− 2 τ + Be− 2 τ .
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Section 4: Solving quadratics
10
4. Solving quadratics
To solve the quadratic equation:
am2 + bm + c = 0,
where a, b, c are constants,
one can sometimes identify simple linear factors that multiply together
to give the left-hand-side of the equation.
Alternatively, one can always use the quadratic formula:
m1,2 =
−b ±
√
b2 − 4ac
,
2a
to find the values of m (denoted m1 and m2 ) that satisfy the quadratic
equation.
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Section 5: Tips on using solutions
11
5. Tips on using solutions
● When looking at the THEORY, ANSWERS, SOLVING QUADRATICS or TIPS pages, use the Back button (at the bottom of the page)
to return to the exercises
● Use the solutions intelligently. For example, they can help you get
started on an exercise, or they can allow you to check whether your
intermediate results are correct
● Try to make less use of the full solutions as you work your way
through the Tutorial
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Solutions to exercises
12
Full worked solutions
Exercise 1. 2y 00 + 3y 0 − 2y = 0
Set y = Aemx →
i.e.
2m2 y + 3my − 2y = 0
i.e.
2m2 + 3m − 2 = 0 :
i.e.
(2m − 1)(m + 2) = 0
i.e.
m1 =
i.e.
1
2
y1 = Ae
dy
= mAemx = my
dx
d2 y
= m2 Aemx = m2 y
dx2
Auxiliary Equation (A.E.)
and m2 = −2 :
1
2x
−2x
and y2 = Be
Two Different Real Roots
: Two Independent Solutions
1
General solution is y = y1 + y2 = Ae 2 x + Be−2x , (A, B are arbitrary
constants).
Return to Exercise 1
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Solutions to exercises
13
Exercise 2. y 00 − 2y 0 + 2y = 0
Set y = Aemx ,
dy
d2 y
= my,
= m2 y
dx
dx2
→
m2 − 2m + 2 = 0
Auxiliary Equation (A.E.)
i.e.
am2 + bm + c = 0
with solutions m =
1
2a
−b ±
√
Since
b2 − 4ac = (−2)2 − 4 · 1 · 2 = 4 − 8 = −4 < 0 ,
In fact,
expect Complex (Conjugate) Solutions.
√ √ m = 21 2 ± −4 = 12 2 ± 2 −1 = 1 ± i .
For m = p ± iq ,
i.e.
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b2 − 4ac .
y = epx (A cos qx + B sin qx)
y = ex (A cos x + B sin x),
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since p = 1 and q = 1 .
Return to Exercise 2
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Solutions to exercises
14
Exercise 3. y 00 − 2y 0 + y = 0
i.e.
m2 − 2m + 1 = 0
(A.E.)
(m − 1)2 = 0
i.e. m = 1 (twice)
i.e.
EQUAL REAL ROOTS
Multiply one solution by x, to get two independent solutions
i.e.
y = y1 + xy2 = Aex + xBex = (A + Bx)ex .
Return to Exercise 3
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Solutions to exercises
15
Exercise 4. y 00 = −4y
√
√
m2 = −4
i.e.
m=
(A.E.)
i.e.
complex conjugate solutions of form p ± iq
with p = 0, q = 2 .
General solution,
−1 ·
4 = i · (±2) = ±2i
y = epx (A cos qx + B sin qx)
i.e.
y = e0 (A cos 2x + B sin 2x)
i.e.
y = A cos 2x + B sin 2x .
Return to Exercise 4
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Solutions to exercises
16
Exercise 5. y 00 = 4y
A.E. is
m2 = 4
i.e.
m = ±2
i.e.
real different roots
∴
y = y1 + y2 = Ae2x + Be−2x .
Return to Exercise 5
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Solutions to exercises
17
Exercise 6. 3y 00 − 36y 0 + 13y = 0
A.E. is
i.e.
i.e.
∴
36m2 − 36m + 13 = 0
p
1
m = 2·36
36 ± (−36)2 − 4 · 36 · 13
=
1
2
=
1
2
±
1±
1
2
q
1−
q
−16
36
± 13 i
1
y = e 2 x A cos
m=
4·13
36
=
1
2
±
i
2
4
6
·
1
2
1
3x
+ B sin
1
3x
.
Return to Exercise 6
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Solutions to exercises
18
Exercise 7. 3y 00 + 2y 0 = 0
A.E. is
3m2 + 2m = 0
Real different roots
:
i.e.
i.e.
m(3m + 2) = 0
i.e.
m1 = 0 and m2 = − 23
2
y = Ae0·x + Be− 3 x
2
y = A + Be− 3 x .
Return to Exercise 7
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Solutions to exercises
19
Exercise 8. 16y 00 − 8y 0 + y = 0
A.E. is
16m2 − 8m + 1 = 0
Real equal roots:
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i.e.
(4m − 1)2 = 0
i.e.
m=
1
4
(twice)
1
y = (A + Bx)e 4 x .
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Solutions to exercises
20
Exercise 9. y 00 + 4y 0 + 5y = 0 ; y(0) = 0
A.E.
m2 + 4m + 5 = 0,
m=
1
2
−4 ±
√
y 0 (0) = 2
and
16 − 20
= −2 ± 22 i = −2 ± i
y = e−2x (A cos x + B sin x) .
General solution is
Particular solution
has
y = 0 when x = 0
i.e.
0 = e0 (A cos(0) + B sin(0))
= 1 · (A + 0) = A
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i.e.
A=0.
And
dy
dx
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= 2 when x = 0
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Solutions to exercises
A = 0 gives
21
y = e−2x B sin x
dy
dx
= −2e−2x B sin x + e−2x B cos x
= Be−2x (cos x − 2 sin x)
i.e.
2 = B · e0 [cos(0) − 2 sin(0)]
= B · 1[1 − 0]
i.e.
B =2.
∴ particular solution is
y = 2e−2x sin x .
Return to Exercise 9
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Solutions to exercises
22
Exercise 10. y 00 + 6y 0 + 13y = 0 ; y(0) = 2
A.E.
m2 + 6m + 13 = 0
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y 0 (0) = 1
√
−6 ± 36 − 52
√
= −3 ± 12 −16
1
2
i.e.
m=
i.e.
m = −3 ± 2i
y = e−3x (A cos 2x + B sin 2x) .
General solution is
Particular solution
and
has
y = 2 when x = 0
i.e.
2 = e0 (A cos(0) + B sin(0))
i.e.
A=2.
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Solutions to exercises
dy
dx
23
= −3e−3x (A cos 2x + B sin 2x) + e−3x (−2A sin 2x + 2B cos 2x)
= e−3x [(2B − 3A) cos 2x − (3B + 2A) sin 2x]
1 = e0 [(2B − 3A) cos(0) − (3B + 2A) sin(0)]
i.e.
i.e.
i.e.
i.e.
∴
dy
dx
= 1 when x = 0
1 = 2B − 3A
1 = 2B − 6 (using A = 2)
7
2 = B.
Particular solution is
y = 12 e−3x (4 cos 2x + 7 sin 2x) .
Return to Exercise 10
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Solutions to exercises
24
Exercise 11. y 00 + 6y 0 + 9y = 0 ; y(0) = 1
A.E. is
m2 + 6m + 9 = 0
i.e.
(m + 3)2 = 0
i.e.
m = −3 (twice)
General solution
:
y = (A + Bx)e−3x .
Particular solution
has
y = 1 when x = 0
i.e.
1 = (A + 0)e0
i.e.
A=1.
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y 0 (0) = 2
and
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Solutions to exercises
dy
dx
25
= Be−3x + (A + Bx) · (−3)e−3x
= e−3x [B − 3(A + Bx)]
i.e.
2 = e0 [B − 3(A + 0)],
i.e.
2 = B − 3A
i.e.
B=5
∴
since
dy
dx
= 2 when x = 0
(using A = 1) .
Particular solution is
y = (1 + 5x)e−3x .
Return to Exercise 11
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Solutions to exercises
Exercise 12.
d2 y
dx2
26
+
dy
dx
+y =0
Set y = Aemx ,
A.E.
i.e.
dy
= my ,
dx
m2 + m + 1 = 0
√
m = 12 [−1 ± 1 − 4] = − 12 ±
d2 y
= m2 y
dx2
√
3
2 i
i.e. p ± iq with
p = − 12
√
q=
General solution:
i.e.
px
3
2
y = e (A cos qx + B sin qx)
√
√ x
y = e− 2 A cos 23x + B sin 23x .
Return to Exercise 12
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Solutions to exercises
Exercise 13.
d2 y
dτ 2
27
dy
− 6 dτ
+ 9y = 0
A.E.
m2 − 6m + 9 = 0
i.e.
(m − 3)2 = 0
i.e.
m=3
and
y = Ae3τ
(twice)
(twice)
To get two independent solutions, multiply one by τ
i.e.
y = (A + Bτ )e3τ .
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Solutions to exercises
Exercise 14.
d2 y
dτ 2
28
dy
+ 7 dτ
+ 12y = 0
A.E.
m2 + 7m + 12 = 0
i.e.
(m + 4)(m + 3) = 0
i.e.
m1 = −4
and
m2 = −3
i.e. two different real roots giving two independent solutions
i.e.
y1 = Ae−4τ
and
y2 = Be−3τ
general solution is y = y1 + y2 = Ae−4τ + Be−3τ .
Return to Exercise 14
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Solutions to exercises
Exercise 15.
d2 x
dτ 2
29
+ 5 dx
dτ + 6x = 0
Set x = Aemτ , i.e.
A.E.
m2 + 5m + 6 = 0
i.e.
(m + 2)(m + 3) = 0
i.e.
m1 = −2
and
dx
= mx,
dτ
m2 = −3
and
d2 x
= m2 x
dτ 2
(two different real roots)
These give two independent solutions x(τ )
i.e.
x1 = Ae−2τ
general solution is
x(τ ) = Ae−2τ + Be−3τ .
and
x2 = Be−3τ
Return to Exercise 15
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Solutions to exercises
30
2
Exercise 16. 4 ddτx2 + 8 dx
dτ + 3x = 0
A.E.
4m2 + 8m + 3 = 0
i.e.
(2m + 3)(2m + 1) = 0
i.e.
m1 = − 32
i.e.
and m2 = − 12
− 32 τ
x1 (τ ) = Ae
(different real roots)
1
and x2 (τ ) = Be− 2 τ
3
(independent solutions)
1
∴ general solution is x(τ ) = Ae− 2 τ + Be− 2 τ .
Return to Exercise 16
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